Molar Mass

[DICK]

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If you have a copy of the Kaplan Blue Book,

If you are smart in Chemistry,

If you are kind and generous, and willing to help others,

Then can you please turn to page 278 and explain the Example on the bottom of that page. Thank you. I am having some trouble with it, I don't know if I understand the concept fully, I was hoping someone could help me out.

And, as always, Thanks Again!

 

[DICK]

If you have a copy of the Kaplan Blue Book,

If you are smart in Chemistry,

If you are kind and generous, and willing to help others,

Then can you please turn to page 278 and explain the Example on the bottom of that page. Thank you. I am having some trouble with it, I don't know if I understand the concept fully, I was hoping someone could help me out.

And, as always, Thanks Again!

no need to satisfy those 3 silly requirements above, here's my question:

What is the molar mass of a 2 L sample of gas that weights 8 g at a temperature of 15 degree Celsius and a pressure of 1.5 atm?

First one to answer correctly AND explains their solution gets a thumbs up icon from me in my next post. Who's gonna get that 'thumbs up' icon. If I don't think you did a good job of explaining your answer (even if you get the correct answer), you get a 'thumbs down'

I also give out two thumbs up ( 👍 👍 ) or two thumbs way up, or two thumbs down .

Sorry to get off tangent, but I would appreciate any explanations,

thanks

 

[dlink]

no need to satisfy those 3 silly requirements above, here's my question:

What is the molar mass of a 2 L sample of gas that weights 8 g at a temperature of 15 degree Celsius and a pressure of 1.5 atm?

First one to answer correctly AND explains their solution gets a thumbs up icon from me in my next post. Who's gonna get that 'thumbs up' icon. If I don't think you did a good job of explaining your answer (even if you get the correct answer), you get a 'thumbs down'

I also give out two thumbs up ( 👍 👍 ) or two thumbs way up, or two thumbs down .

Sorry to get off tangent, but I would appreciate any explanations,

thanks

PV = nRT
PV = (m/M)RT where m = mass of gas
M = molar mass of gas
R = 0.0821 liter.atm/(K.mol)
m = 8 g
V = 2 L
T = 15 + 273 = 288 K
P = 1.5 atm

M = (m/V)[(RT)/P]
= 63.05 g/mol

Note: (m/V) is equivalent to density of the gas

 

[DICK]

PV = nRT
PV = (m/M)RT where m = mass of gas
M = molar mass of gas
R = 0.0821 liter.atm/(K.mol)
m = 8 g
V = 2 L
T = 15 + 273 = 288 K
P = 1.5 atm

M = (m/V)[(RT)/(PV)]
= 63.05 g/mol

Note: (m/V) is equivalent to density of the gas

answer is 63.2, where did you go wrong? your choice value for R? How many decimal places did you use?

BTW, the book I have doesn't even use your method

I think thats why they got the exact answer (no calc needed)

maybe i can try map out what they did, and you or anyone else can explain...

d=8g/2L

V(STP) = 2L (273 K/288 K) (1.5 atm/1 atm) = 2.84 L

8g / 2.84g = 2.82 g/L at STP

(2.82 g / L) (22.4 L / mol) = 63.2 g/mol

I KNOW THIS METHOD IS LONGER, BUT I WOULD LOVE TO KNOW THE REASONING BEHIND IT,

thanks to whomever explains it well enough for me to understand it!

by the way dlink, as promised, 👍 for you efforts, 😉

 

[dat_student]

If you post the question I'd be more than happy to help...I don't have the blue book...I haven't even read the G'Chem section of the white books...May do that the night before my exam...I find Kaplan's G'Chem lessons dull...

 

[DICK]

If you post the question I'd be more than happy to help...I don't have the blue book...I haven't even read the G'Chem section of the white books...May do that the night before my exam...I find Kaplan's G'Chem lessons dull...

Hey, thanks Bud 👍

What is the molar mass of a 2 L sample of gas that weights 8 g at a temperature of 15 degree Celsius and a pressure of 1.5 atm?

 

[puncho]

answer is 63.2, where did you go wrong? your choice value for R? How many decimal places did you use?

BTW, the book I have doesn't even use your method

I think thats why they got the exact answer (no calc needed)

maybe i can try map out what they did, and you or anyone else can explain...

d=8g/2L

V(STP) = 2L (273 K/288 K) (1.5 atm/1 atm) = 2.84 L

8g / 2.84g = 2.82 g/L at STP

(2.82 g / L) (22.4 L / mol) = 63.2 g/mol

I KNOW THIS METHOD IS LONGER, BUT I WOULD LOVE TO KNOW THE REASONING BEHIND IT,

thanks to whomever explains it well enough for me to understand it!

by the way dlink, as promised, 👍 for you efforts, 😉

They're using the fact that P(stp)V(stp)/T(stp) = PV/T

Then, using that they found the V(stp) and the density of the 8g of material at stp.

Now, we know that 1 mol is equal to 22.4L at stp. So, taking the density (which equals m/V), d*V(stp) gives the molar mass.

edit: the other way is also correct...I never did it this way (the 2nd way) I like the PV = nRT better...

 

[dat_student]

Hey, thanks Bud 👍

What is the molar mass of a 2 L sample of gas that weights 8 g at a temperature of 15 degree Celsius and a pressure of 1.5 atm?

Assuming R = 0.08 you approximately get:
MW = mRT / PV = 8 * 0.08 * 288 / 2 * 1.5 = 61.44 (R > 0.08 so the exact number should be slightly higher than 61.44 )

 

[mitshi]

answer is 63.2, where did you go wrong? your choice value for R? How many decimal places did you use?

BTW, the book I have doesn't even use your method

I think thats why they got the exact answer (no calc needed)

maybe i can try map out what they did, and you or anyone else can explain...

d=8g/2L

V(STP) = 2L (273 K/288 K) (1.5 atm/1 atm) = 2.84 L

8g / 2.84g = 2.82 g/L at STP

(2.82 g / L) (22.4 L / mol) = 63.2 g/mol

I KNOW THIS METHOD IS LONGER, BUT I WOULD LOVE TO KNOW THE REASONING BEHIND IT,

thanks to whomever explains it well enough for me to understand it!

by the way dlink, as promised, 👍 for you efforts, 😉

You should have combined the last two calculations and make your final answer 63.10. This method actually gives rise to greater error as in the 3 calculations you show here, 3 truncations have been made. Anyway, these odd figures used by Kaplan will not be relective of what you're going to see in the DAT. 😉